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The combustion of liquid ethanol (C2H5OH) produces carbon

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 92P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 92P

PROBLEM 92P

The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 3.8 mL of ethanol (density = 0.789 g/mL) is allowed to burn in the presence of 12.5 g of oxygen gas, 3.10 mL of water (density = 1.00 g/mL) is collected. Determine the limiting reactant, theoretical yield of H2O, and percent yield for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

Step-by-Step Solution:
Step 3 of 5

Solution 92P

Step1:

C2H5OH + 3O2 → 2CO2+ 3H2O

Given that initial volume of ethanol is 3.8 mL and density is 0.789 g/mL

Mass= density volume=3.8 mL0.789 g/mL=2.99 g

Number of moles of ethanol = Given mass/Molecular mass

Molecular mass of ethanol=46g/mol

Thus number of moles of ethanol==0.06 moles

Step 2:

Given that mass of oxygen is 12.5 g

Number of moles of oxygen =Given mass/Molecular mass

Molecular mass of oxygen=32g/mol

Thus number of moles of oxygen==0.390 moles

Step 3:

From the equation it is clear that 1 mole of ethanol combines with 3 mole of  oxygen to give 2 moles carbon dioxide and 3 moles of water.Thus we can say that the ratio in which both the reactants should be present is 1:3.Since lesser amount of ethanol (less than 0.13 which is 1/3rd of 0.390)is present than required,it should be the limiting reagent.

Step 4 of 5

Chapter 8, Problem 92P is Solved
Step 5 of 5

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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